Problem: Multiply the following complex numbers: $({4-i}) \cdot ({1+3i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4-i}) \cdot ({1+3i}) = $ $ ({4} \cdot {1}) + ({4} \cdot {3}i) + ({-1}i \cdot {1}) + ({-1}i \cdot {3}i) $ Then simplify the terms: $ (4) + (12i) + (-1i) + (-3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 4 + (12 - 1)i - 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 4 + (12 - 1)i - (-3) $ The result is simplified: $ (4 + 3) + (11i) = 7+11i $